Below is my question. Q7.9 is what I'm stuck on. I've done Q7.8; I included it in the picture because I'll use it in Q7.9, and it gives a definition that I'll use.
Update: This question is now solved, and I've added the details below.
What I've done so far is this:
By using time-inversion, $(tB_{1/t})_{t\ge0}$, and $(-B_t)_{t\ge0}$, sign-inversion of Brownian motion, we have that the four random variables in question are all equal to each other. Further, we see that the first is $\mathcal{F}_{0^+}$ measurable, since
$$ \limsup_{t \to 0} \frac{B_t}{\sqrt{t}} = \lim_{s \to 0} \sup_{t \le s} \frac{B_s}{\sqrt{s}}, \ \text{and} \ \sup_{t \le s} \frac{B_s}{\sqrt{s}}$$
is $\mathcal{F}_s$ measurable for all $s > 0$, and thus the limit is $\mathcal{F}_{0^+}$ measurable. Hence by Blumenthal's $0$-$1$ law, it is almost surely constant. Hence we now have that the four random variables in question are equal to each other and almost surely constant. Since $\mathbb{P}(B_{t'} > 0) = 1/2$ for all $t' > 0$, by the Markov property, we have that the almost sure constant must be at least $0$.
I'm stuck on showing that this constant is in fact $+\infty$. I wanted to use the scaling property, $(cB_{t/c^2})_{t\ge0}$, of Brownian motion, but the issue is that this gives
$$\frac{cB_{t/c^2}}{\sqrt{t}} = \frac{B_{t/c^2}}{\sqrt{t/c^2}} = \frac{B_s}{\sqrt{s}}$$
where $s = t/c^2$. When considering just $B_t$ or $B_t/t$, we get a factor $c$ or $1/c$ out the front, and so, since this must hold for all $c$, we know that it must be $0$ or $\infty$. (We can then show which it is.) However, we don't get this nice property when using $B_t/\sqrt{t}$.
Solution: Define $A^x_t$ and $A^x$ as follows: $$A^x_t = \left\{ \sup_{s \le t} \frac{B_s}{\sqrt{s}} \le x \right\}, ~ A^x = \left\{ \lim_{t \downarrow 0}\sup_{s \le t} \frac{B_s}{\sqrt{s}} \le x \right\} = \left\{ \limsup_{t \downarrow 0} \frac{B_s}{\sqrt{s}} \le x \right\}.$$ Observe that $B_t/\sqrt{t} \sim N(0,1)$. Thus, since $\sup_{s \le t} {B_s}/{\sqrt{s}} \ge {B_t}/{\sqrt{t}}$, $$P \left( \sup_{s \le t} \frac{B_s}{\sqrt{s}} \le x \right) \le P \left( \frac{B_t}{\sqrt{t}} \le x \right) = \Phi(x),$$ where $\Phi$ is the cdf for the standard normal. We want to show that $P(A^x) = 0$ for every $x \in \mathbb{R}$ ($\therefore x \neq \infty$); by Blumenthal's $0$-$1$ law, it is enough to show that $P(A^x) < 1$. Now, $A^x_{1/n} \downarrow A^x$ as $n \to \infty$, so by monotone convergence, $$P(A^x) = \lim_{n \to \infty}P(A^x_{1/n}) \le \Phi(x) < 1, \ x \in \Bbb R.$$ Thus $P(A^x) < 1$, ie $P(A^x) = 0$, for all $x \in \mathbb{R}$. Thus the almost sure constant must be $+\infty$.
Thank you to Jay.H for helping me with this!
Try to use the fact that $B_t/\sqrt{t}$ has the same distribution as $B_1$ and the fact that $\sup_{s\le t }B_s/\sqrt{s}$ is monotonic w.r.t. $t$
[Warning: more details below]
For any constant $C>0$, let $$ A = (\lim_{t\to 0} \sup_{0<s\le t}\frac{B_s}{\sqrt{s}}<C )$$ We want to show that $P(A)=0$. Since $A\in {\mathcal F}_{0+}$, we only have to show that $P(A)<1$.
Let $$ A_n = (\sup_{0<s\le \frac{1}{n}}\frac{B_s}{\sqrt{s}}<C)$$ Then $$ P(A_n) \le P(\frac{B_{1/n}}{\sqrt{1/n}}<C) = P(B_1<C)$$ Since $ A_n \uparrow A$, we have $$ P(A) = \lim_{n\to \infty} P(A_n) \le P(B_1<C) <1$$ Done.